Metamath Proof Explorer


Theorem srgisid

Description: In a semiring, the only left-absorbing element is the additive identity. Remark in Golan p. 1. (Contributed by Thierry Arnoux, 1-May-2018)

Ref Expression
Hypotheses srgz.b B = Base R
srgz.t · ˙ = R
srgz.z 0 ˙ = 0 R
srgisid.1 φ R SRing
srgisid.2 φ Z B
srgisid.3 φ x B Z · ˙ x = Z
Assertion srgisid φ Z = 0 ˙

Proof

Step Hyp Ref Expression
1 srgz.b B = Base R
2 srgz.t · ˙ = R
3 srgz.z 0 ˙ = 0 R
4 srgisid.1 φ R SRing
5 srgisid.2 φ Z B
6 srgisid.3 φ x B Z · ˙ x = Z
7 6 ralrimiva φ x B Z · ˙ x = Z
8 1 3 srg0cl R SRing 0 ˙ B
9 oveq2 x = 0 ˙ Z · ˙ x = Z · ˙ 0 ˙
10 9 eqeq1d x = 0 ˙ Z · ˙ x = Z Z · ˙ 0 ˙ = Z
11 10 rspcv 0 ˙ B x B Z · ˙ x = Z Z · ˙ 0 ˙ = Z
12 4 8 11 3syl φ x B Z · ˙ x = Z Z · ˙ 0 ˙ = Z
13 7 12 mpd φ Z · ˙ 0 ˙ = Z
14 1 2 3 srgrz R SRing Z B Z · ˙ 0 ˙ = 0 ˙
15 4 5 14 syl2anc φ Z · ˙ 0 ˙ = 0 ˙
16 13 15 eqtr3d φ Z = 0 ˙